In a system A(s)⇌B(g)+3C(g), if the equilibrium concentration ofB increases by a factor of 8, it will cause the equilibrium concentration of C to change to

In a system $A (s) ⇌ B (g) + 3 C (g)$ , if the equilibrium concentration ofB increases by a factor of 8, it will cause the equilibrium concentration of C to change to

A
two times its original value
B
one half of its original value
C
eight times its original value
D
one-eight its original value

Solution:

$[B]_{1} [C]_{1}^{3} = [B]_{2} [C]_{2}^{3}$

It is given that $[B]_{2} = 8 [B]_{1} .$ Hence $[C]_{2} = \frac{[C]_{1}}{8^{1 / 3}} = \frac{[C]_{1}}{2}$

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